Exercises — Gradient perpendicular to level curves - surfaces

$f_x=2x$, $f_y=6y$. So $\nabla f=(2x,6y)$, and at $(1,2)$ that is $==(2,12)==$. The level curve $f=13$ is an ellipse. The gradient points outward across it, in the direction $f$ increases (you climb as you leave the origin). It is perpendicular to the ellipse's tangent at $(1,2)$.

$\nabla f=(2x,2y,2z)===(2,4,6)==$ at $(1,2,3)$. Because $\nabla f$ is perpendicular to the level surface (a sphere of radius $\sqrt{14}$), this vector is a normal. Note it equals $2(1,2,3)$ — radially outward, as expected for a sphere.

$\nabla f=(y,x)=(1,4)$ at $(4,1)$. The tangent must dot to $0$ with $\nabla f$.

• $(1,4)\cdot (1,4)=1+16=17\ne 0$ — this is the normal, not the tangent.
• $(-1,4)\cdot (1,4)=-1+16=15\ne 0$ — also not tangent! Neither given option is tangent; the true tangent is perpendicular to $(1,4)$, e.g. $==(-4,1)==$ (since $(1,4)\cdot (-4,1)=-4+4=0$). This is a trap-spotting drill: don't accept an option just because it "looks sideways."

Let $f=x^2+4y^2$, so $\nabla f=(2x,8y)=(4,8)$ at $(2,1)$. This is the normal. Point–normal form: $\nabla f\cdot ((x,y)-(2,1))=0$: $4(x-2)+8(y-1)=0\Rightarrow 4x+8y=16\Rightarrow ==x+2y=4==.$ Check: $(2,1)$ gives $2+2=4$. ✅

$\nabla f=(2x,2y,2z)=(2,-2,4)$ at $(1,-1,2)$. $2(x-1)-2(y+1)+4(z-2)=0\Rightarrow 2x-2y+4z=12\Rightarrow ==x-y+2z=6==.$ Check: $1-(-1)+2(2)=1+1+4=6$. ✅

Write the surface as a level set $g=0$ with $g=x^2+y^2-z$, so we can use the gradient-normal trick. $\nabla g=(2x,2y,-1)=(2,2,-1)$ at $(1,1,2)$. $2(x-1)+2(y-1)-1(z-2)=0\Rightarrow ==2x+2y-z=2==.$ Check: $2+2-2=2$. ✅ (Notice the $-1$ in the third slot — that is exactly the graph-normal $(f_x,f_y,-1)$ warned about in the parent note.)

Gradient way. $f=xy$, $\nabla f=(y,x)=(2,3)$ at $(3,2)$. Tangent is perpendicular to $(2,3)$, e.g. $\mathbf{t}=(-3,2)$ (since $(2,3)\cdot (-3,2)=-6+6=0$). Slope $=\frac{2}{-3}=-\frac{2}{3}$. Implicit-slope way. $\frac{dy}{dx}=-\frac{f_x}{f_y}=-\frac{y}{x}=-\frac{2}{3}=-\frac{2}{3}$. Both give slope $==-\frac{2}{3}==$. ✅ The look-at-the-figure picture: the red $\nabla f$ arrow spears the curve at a right angle; the blue tangent runs along it.

$\nabla f=(2x,2y)=(6,8)$. The tangent to the circle at $(3,4)$ is perpendicular to the radius, direction $(-4,3)$; as a unit vector $\mathbf{u}=\frac{1}{5}(-4,3)$. $D_{\mathbf{u}}f=\nabla f\cdot \mathbf{u}=(6,8)\cdot \frac{1}{5}(-4,3)=\frac{1}{5}(-24+24)===0==.$ It's zero because moving along a level curve keeps $f$ constant — the Directional Derivative in the tangent direction must vanish. This is the parent result seen as a number.

$\nabla f=(2x,2y)$ is parallel to $(x,y)$. We want $(x,y)\parallel (1,1)$, so $x=y$, with $x^2+y^2=25$: $2x^2=25$, $x=\frac{5}{\sqrt{2}}$. Point $==\left(\frac{5}{\sqrt{2}},\frac{5}{\sqrt{2}}\right)==$ (the outward-pointing one). There $\nabla f=\frac{10}{\sqrt{2}}(1,1)$, indeed along $(1,1)$.

The intersection curve lies in both surfaces, so its tangent is perpendicular to both normals. Compute the normals:

• $\nabla f=(2x,2y,2z)=(2,2,4)$.
• $\nabla g=(2x,2y,-1)=(2,2,-1)$. A vector perpendicular to both is the cross product $\nabla f\times \nabla g$ (the cross product is the tool that hands you a vector orthogonal to two given ones): $\nabla f\times \nabla g=\left|\begin{array}{ccc}\mathbf{i}& \mathbf{j}& \mathbf{k}\\ 2& 2& 4\\ 2& 2& -1\end{array}\right|=(2(-1)-4(2),4(2)-2(-1),2(2)-2(2))=(-10,10,0).$ Tangent direction $==(-10,10,0)==$, or simplified $(-1,1,0)$. Check both dots are zero: $(2,2,4)\cdot (-1,1,0)=0$ and $(2,2,-1)\cdot (-1,1,0)=0$. ✅

$f=\frac{x^2}{4}+y^2$, $\nabla f=\left(\frac{x}{2},2y\right)$. We need $\nabla f\parallel (1,1)$: $\frac{x}{2}=2y\Rightarrow x=4y$. Substitute into the ellipse: $\frac{(4y{)}^2}{4}+y^2=1\Rightarrow 4y^2+y^2=1\Rightarrow y^2=\frac{1}{5}$. First quadrant: $y=\frac{1}{\sqrt{5}}$, $x=\frac{4}{\sqrt{5}}$. Point $==\left(\frac{4}{\sqrt{5}},\frac{1}{\sqrt{5}}\right)==$.

Steepest decrease is $-\nabla f$. Here $\nabla f=(6,8)$, magnitude $10$, so the unit descent direction is $-\frac{1}{10}(6,8)=(-0.6,-0.8)$. New point $=(3,4)+1\cdot (-0.6,-0.8)=(2.4,3.2)$. Landing point $==(2.4,3.2)==$. It moves straight toward the origin (perpendicular to the level circle), because the gradient is radial here.

Let $\mathbf{r}(t)$ be any smooth curve on the surface with $\mathbf{r}(0)={\mathbf{x}}_0$, so $f(\mathbf{r}(t))=c$ for all $t$. Differentiate with the Multivariable Chain Rule: $\frac{d}{dt}f(\mathbf{r}(t))=\nabla f(\mathbf{r}(t))\cdot {\mathbf{r}}^{\prime }(t)=\frac{d}{dt}c=0.$ At $t=0$: $\nabla f({\mathbf{x}}_0)\cdot {\mathbf{r}}^{\prime }(0)=0$. Every tangent vector ${\mathbf{r}}^{\prime }(0)$ arises from such a curve, so $\nabla f({\mathbf{x}}_0)$ is orthogonal to all of them — it is the normal. The tangent plane is the set of $\mathbf{x}$ with $\mathbf{x}-{\mathbf{x}}_0$ perpendicular to that normal: $\nabla f({\mathbf{x}}_0)\cdot (\mathbf{x}-{\mathbf{x}}_0)=0$. $■$

$f=0$ means $x^2=y^2$, i.e. the two crossing lines $y=x$ and $y=-x$. $\nabla f=(2x,-2y)=(0,0)$ at the origin — the gradient vanishes. When $\nabla f=0$ the theorem's hypothesis fails: there is no well-defined normal direction, and indeed the level set has a crossing point (two tangent lines, not one). This is exactly the singular point case: "perpendicular to a unique tangent" only makes sense where $\nabla f\ne 0$.

Graph normal. Write the graph as a level set $g(x,y,z)=f(x,y)-z=x^2+y^2-z=0$. Then $\nabla g=(f_x,f_y,-1)=(2,2,-1)$. That $3$-vector $==(2,2,-1)==$ is normal to the graph in $3$D. Level-curve normal. In the $2$D domain, $\nabla f=(2x,2y)=(2,2)$ at $(1,1)$ is normal to the level curve $x^2+y^2=2$. They are different objects: $(2,2)$ lives in the flat input plane and points across the contour; $(2,2,-1)$ lives in $3$D and tilts off the graph. The first two entries match because the graph normal's horizontal shadow is the level-curve normal — a satisfying consistency, not a coincidence.

Perpendicularity comes from which identity? ::: $\nabla f(\mathbf{r}(t))\cdot {\mathbf{r}}^{\prime }(t)=0$ (chain rule on $f=c$). Tangent to a curve of intersection of two surfaces? ::: $\nabla f\times \nabla g$ (perpendicular to both normals). Steepest descent direction? ::: $-\nabla f$, perpendicular to the level curve. When does the whole result break? ::: When $\nabla f=0$ (singular point, no unique tangent).